Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

If x and y are positive integers such that x=8y+12, what is [#permalink]
05 Jun 2005, 17:43

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

14% (00:00) correct
86% (00:59) wrong based on 8 sessions

If x and y are positive integers such that x=8y+12, what is the greatest common divisor of x and y?
1) x=12u, where u is an integer.
2) y=12z, where z is an integer.

If x and y are positive integers such that x=8y+12, what is the greatest common divisor of x and y? 1) x=12u, where u is an integer. 2) y=12z, where z is an integer.

1) y = (x - 12)/8 = 12(u-1)/8 and 12u so u-1 must be divisible by 2, 12 may not be a GCD, the GDC for these two numbers is at least 2, but depending on u it can be greater than 2.
say u-1 is divisible by 8 then GDC can be 12.
so insufficient

2) y=12z and x = 12(8z+1), here GCD is 12, is there a GCD >1 for z and 8z+1? no because
suppose there is a divisor m such that z/m = integer then 8z/m + 1/m is not.

If x and y are positive integers such that x=8y+12, what is the greatest common divisor of x and y? 1) x=12u, where u is an integer. 2) y=12z, where z is an integer.

1) y = (x - 12)/8 = 12(u-1)/8 and 12u so u-1 must be divisible by 2, 12 may not be a GCD, the GDC for these two numbers is at least 2, but depending on u it can be greater than 2. say u-1 is divisible by 8 then GDC can be 12. so insufficient

2) y=12z and x = 12(8z+1), here GCD is 12, is there a GCD >1 for z and 8z+1? no because suppose there is a divisor m such that z/m = integer then 8z/m + 1/m is not.

so GDC is 12 and sufficient. B.

sparky, agree with you. but more than this you have nice approach to handle the problem. thanx. keep up the sprit.

If x and y are positive integers such that x=8y+12, what is the greatest common divisor of x and y? 1) x=12u, where u is an integer. 2) y=12z, where z is an integer.

1) y = (x - 12)/8 = 12(u-1)/8 and 12u so u-1 must be divisible by 2, 12 may not be a GCD, the GDC for these two numbers is at least 2, but depending on u it can be greater than 2. say u-1 is divisible by 8 then GDC can be 12. so insufficient

2) y=12z and x = 12(8z+1), here GCD is 12, is there a GCD >1 for z and 8z+1? no because suppose there is a divisor m such that z/m = integer then 8z/m + 1/m is not.

so GDC is 12 and sufficient.

B.

i do not understand why 12 cant be gcd of statement 1 ... Can someone please explain ??? and I dont understand the reasoning behind

12u=8y+12; y=3/2(u-1). Given y is a positive integer, so u=3,5,7...; y=3,6,9... For y=3; x=36. GCD(3,36) is 3. For y=6; x=60. GCD(6,60) is 6. Insufficient

Statement 2 y=12z x=8*12z+12 or 12(8z+1). Both x and y are multiples of 12. Therefore GCD is 12. Sufficient